# A triangle has corners at (-4 ,2 ), (7 ,1 ), and (2 ,-7 ). If the triangle is dilated by a factor of 2/5  about point #(-1 ,4 ), how far will its centroid move?

the centroid will move by about $\frac{8 \sqrt{5}}{5}$
$\Delta = \frac{8 \sqrt{5}}{5}$

$\Delta = 3.5771 \text{ }$units

#### Explanation:

the solution:

the centroid of the given triangle is at
${x}_{o} = \frac{- 4 + 7 + 2}{3} = \frac{5}{3}$

${y}_{o} = \frac{2 + 1 - 7}{3} = - \frac{4}{3}$

the reference point is at $\left(- 1 , 4\right) \text{ }$and scale factor $k = \frac{2}{5}$

Let the unknown new centroid be at $C \left({x}_{c} , {y}_{c}\right)$

By segment division method

$\frac{{x}_{c} - \left(- 1\right)}{{x}_{o} - \left(- 1\right)} = \frac{2}{5}$

$\frac{{x}_{c} - \left(- 1\right)}{\frac{5}{3} - \left(- 1\right)} = \frac{2}{5}$

${x}_{c} = \frac{1}{15}$

Also

$\frac{{y}_{c} - 4}{{y}_{o} - 4} = \frac{2}{5}$

$\frac{{y}_{c} - 4}{- \frac{4}{3} - 4} = \frac{2}{5}$

${y}_{c} = \frac{28}{15}$

The old centroid is at $\left(\frac{5}{3} , - \frac{4}{3}\right) \text{ }$(black colored dot)
The new centroid is at $C \left(\frac{1}{15} , \frac{28}{15}\right) \text{ }$(orange colored dot)

how far did the centroid move ?

$\Delta = \sqrt{{\left({x}_{o} - {x}_{c}\right)}^{2} + {\left({y}_{o} - {y}_{c}\right)}^{2}}$

$\Delta = \sqrt{{\left(\frac{5}{3} - \frac{1}{15}\right)}^{2} + {\left(- \frac{4}{3} - \frac{28}{15}\right)}^{2}}$

$\Delta = \sqrt{\frac{64 + 256}{25}}$

$\Delta = \frac{8 \sqrt{5}}{5}$

$\Delta = 3.5771 \text{ }$units (length of the green segment)

God bless.... I hope the explanation is useful.